Question

The 2nd, 31st and last term of an A.P. is, 7(3/4), 1/2 and -6(1/2) respectively. The number of terms of the A.P. is
(a) 48
(b) 60
(c) 52
(d) 59​

Answer 1

Step-by-step explanation:

तरस गये है हम तेरे मुंह से कुछ सुनने को हम प्यार की बात न सही कोई शिकायत ही कर दे

Answer 2

Answer:

The correct option is (d) i.e 59

Step-by-step explanation:

$\bf \underline{Solution-}$

Let assume that the first term and common difference is a and d respectively.

$\rm \Rightarrow \: a + d = 7 \frac{3}{4}$

$\rm \Rightarrow \:a + d = \frac{31}{4}$

$\rm \Rightarrow \:a + (31 - 1)d = \frac{1}{2}$

$\rm \Rightarrow \:a + 30d = \frac{1}{2}$

Solve the above equation

$\rm \Rightarrow \:d = - \frac{1}{4}$

$\rm \Rightarrow \:a = 8$

Assume that the total number of terms are n

$\rm \Rightarrow \:8 + (n - 1) \bigg( - \frac{1}{4} \bigg) = - \frac{13}{2}$

$\rm \Rightarrow \:8 - \frac{1}{4} n + \frac{1}{4} = - \frac{13}{2}$

$\rm \Rightarrow \: \frac{33}{4} - \frac{1}{4} n = - \frac{13}{2 }$

$\rm \Rightarrow \: \frac{1}{4} n = \frac{33}{4} + \frac{13}{2}$

Solve further

$\rm \Rightarrow \: \frac{1}{4} n = \frac{(33 \times 1) + (13 \times 2)}{4}$

$\rm \Rightarrow \: \frac{1}{4} n = \frac{33 + 26}{4}$

$\rm \Rightarrow \: \frac{1}{4} n = \frac{59}{4}$

$\bf \Rightarrow \:n = 59.$

The correct option is (d) i.e 59.